alexpavlos wrote:

If 4 points are indicated on a line and 5 points are indicated on another line that is parallel to the first line, how many triangles can be formed whose vertices are among the 9 points?

A. 20

B. 30

C. 40

D. 70

E. 90

Tried doing a few options, but couldn't get it right and the answer provided left me more confused!

Approach #1:There are two types of triangles possible:

With two vertices on the line with 4 points and the third vertex on the line with 5 points --> \(C^2_4*C^1_5=30\);

With two vertices on the line with 5 points and the third vertex on the line with 4 points --> \(C^2_5*C^1_4=40\);

Total: \(30+40=70\).

Answer: D.

Approach #2:All

different 3 points out of total 4+5=9 points will create a triangle EXCEPT those 3 points which are collinear.

\(C^3_{9}-(C^3_4+C^3_5)=84-(4+10)=70\) (where \(C^3_4\) and \(C^3_4\) are # of different 3 collinear points possible from the line with 4 points and the line with 5 points, respectively).

Answer: D.

Hope it's clear.